在Python中,格式化字符串时,我可以按名称而不是按位置填充占位符,如下所示:
print "There's an incorrect value '%(value)s' in column # %(column)d" % \
{ 'value': x, 'column': y }
我想知道这在Java中是否可能(希望没有外部库)?
在Python中,格式化字符串时,我可以按名称而不是按位置填充占位符,如下所示:
print "There's an incorrect value '%(value)s' in column # %(column)d" % \
{ 'value': x, 'column': y }
我想知道这在Java中是否可能(希望没有外部库)?
当前回答
不完全是,但你可以使用MessageFormat多次引用一个值:
MessageFormat.format("There's an incorrect value \"{0}\" in column # {1}", x, y);
上面的事情也可以用string .format()来完成,但是如果你需要构建复杂的表达式,我发现messageFormat语法更干净,而且你不需要关心你放入字符串中的对象的类型
其他回答
我最终得到了下一个解决方案: 使用substitute()方法创建类templatessubstitute,并使用它格式化输出 然后创建一个字符串模板,并用值填充它
import java.util.*;
public class MyClass {
public static void main(String args[]) {
String template = "WRR = {WRR}, SRR = {SRR}\n" +
"char_F1 = {char_F1}, word_F1 = {word_F1}\n";
Map<String, Object> values = new HashMap<>();
values.put("WRR", 99.9);
values.put("SRR", 99.8);
values.put("char_F1", 80);
values.put("word_F1", 70);
String message = TemplateSubstitutor.substitute(values, template);
System.out.println(message);
}
}
class TemplateSubstitutor {
public static String substitute(Map<String, Object> map, String input_str) {
String output_str = input_str;
for (Map.Entry<String, Object> entry : map.entrySet()) {
String key = entry.getKey();
Object value = entry.getValue();
output_str = output_str.replace("{" + key + "}", String.valueOf(value));
}
return output_str;
}
}
这是一个旧的线程,但只是为了记录,你也可以使用Java 8风格,像这样:
public static String replaceParams(Map<String, String> hashMap, String template) {
return hashMap.entrySet().stream().reduce(template, (s, e) -> s.replace("%(" + e.getKey() + ")", e.getValue()),
(s, s2) -> s);
}
用法:
public static void main(String[] args) {
final HashMap<String, String> hashMap = new HashMap<String, String>() {
{
put("foo", "foo1");
put("bar", "bar1");
put("car", "BMW");
put("truck", "MAN");
}
};
String res = replaceParams(hashMap, "This is '%(foo)' and '%(foo)', but also '%(bar)' '%(bar)' indeed.");
System.out.println(res);
System.out.println(replaceParams(hashMap, "This is '%(car)' and '%(foo)', but also '%(bar)' '%(bar)' indeed."));
System.out.println(replaceParams(hashMap, "This is '%(car)' and '%(truck)', but also '%(foo)' '%(bar)' + '%(truck)' indeed."));
}
输出将是:
This is 'foo1' and 'foo1', but also 'bar1' 'bar1' indeed.
This is 'BMW' and 'foo1', but also 'bar1' 'bar1' indeed.
This is 'BMW' and 'MAN', but also 'foo1' 'bar1' + 'MAN' indeed.
可以使用Apache Commons StringSubstitutor。注意,StrSubstitutor已弃用。
import org.apache.commons.text.StringSubstitutor;
// ...
Map<String, String> values = new HashMap<>();
values.put("animal", "quick brown fox");
values.put("target", "lazy dog");
StringSubstitutor sub = new StringSubstitutor(values);
String result = sub.replace("The ${animal} jumped over the ${target}.");
// "The quick brown fox jumped over the lazy dog."
这个类支持为变量提供默认值。
String result = sub.replace("The number is ${undefined.property:-42}.");
// "The number is 42."
要使用递归变量替换,调用setEnableSubstitutionInVariables(true);。
Map<String, String> values = new HashMap<>();
values.put("b", "c");
values.put("ac", "Test");
StringSubstitutor sub = new StringSubstitutor(values);
sub.setEnableSubstitutionInVariables(true);
String result = sub.replace("${a${b}}");
// "Test"
在编写本文时,Java中还没有内置任何东西。我建议编写自己的实现。我的偏好是一个简单流畅的构建器接口,而不是创建一个映射并将其传递给函数——你最终会得到一个漂亮的连续代码块,例如:
String result = new TemplatedStringBuilder("My name is {{name}} and I from {{town}}")
.replace("name", "John Doe")
.replace("town", "Sydney")
.finish();
下面是一个简单的实现:
class TemplatedStringBuilder {
private final static String TEMPLATE_START_TOKEN = "{{";
private final static String TEMPLATE_CLOSE_TOKEN = "}}";
private final String template;
private final Map<String, String> parameters = new HashMap<>();
public TemplatedStringBuilder(String template) {
if (template == null) throw new NullPointerException();
this.template = template;
}
public TemplatedStringBuilder replace(String key, String value){
parameters.put(key, value);
return this;
}
public String finish(){
StringBuilder result = new StringBuilder();
int startIndex = 0;
while (startIndex < template.length()){
int openIndex = template.indexOf(TEMPLATE_START_TOKEN, startIndex);
if (openIndex < 0){
result.append(template.substring(startIndex));
break;
}
int closeIndex = template.indexOf(TEMPLATE_CLOSE_TOKEN, openIndex);
if(closeIndex < 0){
result.append(template.substring(startIndex));
break;
}
String key = template.substring(openIndex + TEMPLATE_START_TOKEN.length(), closeIndex);
if (!parameters.containsKey(key)) throw new RuntimeException("missing value for key: " + key);
result.append(template.substring(startIndex, openIndex));
result.append(parameters.get(key));
startIndex = closeIndex + TEMPLATE_CLOSE_TOKEN.length();
}
return result.toString();
}
}
谢谢你的帮助!使用所有的线索,我写了一个例程来做我想要的——使用字典的类似python的字符串格式化。因为我是Java新手,任何提示都是感激的。
public static String dictFormat(String format, Hashtable<String, Object> values) {
StringBuilder convFormat = new StringBuilder(format);
Enumeration<String> keys = values.keys();
ArrayList valueList = new ArrayList();
int currentPos = 1;
while (keys.hasMoreElements()) {
String key = keys.nextElement(),
formatKey = "%(" + key + ")",
formatPos = "%" + Integer.toString(currentPos) + "$";
int index = -1;
while ((index = convFormat.indexOf(formatKey, index)) != -1) {
convFormat.replace(index, index + formatKey.length(), formatPos);
index += formatPos.length();
}
valueList.add(values.get(key));
++currentPos;
}
return String.format(convFormat.toString(), valueList.toArray());
}